#### Answer

$x=0$ or $x=-2$

#### Work Step by Step

We solve by factoring:
$x^{5}+8x^{2}=0$
$x^{2}(x^{3}+8)=0$
$x^{2}(x+2)(x^{2}-2x+4)=0$
$x^{2}=0$ or $x+2=0$ or $x^{2}-2x+4=0$
$x=0$ or $x=-2$
(The last equation $x^{2}-2x+4=0$ has no real solutions because the discriminant is negative.)